Answer:
Step-by-step explanation:
Given
Maximum height H = 300m
Range (horizontal distance) = 380m
Required
Initial speed U and the angle of the ball when it was launched.
Range = U√2H/g
380 = U√2(300)/9.8
380 = U√600/9.8
380 = 7.8246U
U = 380/7.8246
U = 48.57m/s
The initial speed is 48.57m/s
b) Using the formula for calculating time of flight;
T = 2Usin theta/g
9 = 2(48.57)sin theta/9.8
9*9.8 = 97.14sin theta
88.2 = 97.14sin theta
88.2/97.14 = sin theta
sin theta = 0.9079
theta = sin^-1(0.9079)
theta = 65.23°
hence the angle when the ball was launched is 65.23°